You don't have to know any music theory
to do this question as all you need to know is explained here.
You only need to know about logarithms. The actual problems to
solve are given in bold italic text.
Diagram showing part of a piano keyboard.
Wherever you are on the keyboard, the interval between one note
and the next with the same letter name is always an octave. In
the same way, the interval between C and the G above it, or D
and the A above it, or anything similar (seven steps on the
keyboard, including both black and white notes), is always a
fifth. Each of these intervals, the octave, the fifth, and
others, corresponds to a particular ratio of string lengths
which produce the notes. The octave corresponds to 2/1, halving
the length of the string. The fifth corresponds to 3/2, taking
two thirds of the length of the string. The interval from C to
F, called a fourth, has the ratio 4/3.'Fourth' and 'fifth' etc.
are musical terms and do not refer to the fractions 1/4 and
1/5.
The interval from F to G, between the fourth and the fifth, has
the ratio ${3/2 \over 4/3} = 9/8$. This interval is called a
tone and it is also the the interval from C to D. Notes tuned
in these ratios produce pleasant harmonies but compromises have
to be made in tuning the other notes because the ratios do not
relate to each other exactly. This problem is about discovering
what the compromises might be. The table below shows some
musical intervals and the corresponding ratios of the notes
with respect to the note C.
| C |
D |
E |
F |
G |
A |
B |
C |
|
1/1
unison
|
9/8
tone
|
81/64 |
4/3
fourth
|
3/2
fifth
|
27/16 |
243/128 |
2/1
octave
|
The Pythagorean Scale
You might expect there to be three tones in the interval from C
to F, changing the ratio by 9/8 then 9/8 again and then 9/8 a
third time, that is ${(9/8)}^3$.
How does this compare
with the ratio 4/3?
Pythagorean tuning kept
the ideal ratios for the octaves, fifths and fourths and tuned
the note E with a ratio ${(9/8)}^2$ and the notes A and B with
the ratios $(3/2)\times (9/8)$ and $(3/2)\times {(9/8)}^2$. This
system preserves the ratios of the fourth and fifth but produces
a major third, from C to E and a major sixth, from C to A, which
sound unpleasant. The Greeks were not interested in these
intervals, but when composers in the middle ages and Renaissance
wanted to use them, a new system of tuning was needed.
Just intonation (shown in
the table below) has much better major thirds and major sixths,
of 5/4 and 5/3. It tunes the B so that the interval from E to B
is a perfect fifth and the interval from G to B is a perfect
third. But it does this by having two different sizes of tone:
the intervals C-D and D-E are different here, but they were the
same in the Pythagorean scale.
Find the ratio corresponding to B. Also
compare the D-A ratio with the ideal fifth (3/2).
| C |
D |
E |
F |
G |
A |
B |
C |
|
1/1
unison
|
9/8
tone
|
5/4
third
|
4/3
fourth
|
3/2
fifth
|
5/3
sixth
|
? |
2/1
octave
|
Just Intonation
You might expect there to be six tones and twelve semitones in an
octave (because there are 12 notes) but again that is not exactly
so.
Find exactly how many tones
there are in an octave by finding what power of $9\over 8$ gives
2. Find also exactly how many major thirds there are in an
octave.
In the
equal tempered
scale , the standard tuning nowadays, these mismatches
between different ratios are removed by defining the intervals
differently. The tone is made a little smaller, so that there are
exactly six of them in an octave. The semitone (half a tone) is a
twelfth of an octave: so its ratio is ${(2)}^{1/12}$. If, for
example, the note C is tuned with a string length of 32 units
then the C# is tuned with a string of length ${32\over
{(2)}^{1/12}}=30.2$ (to 3 significant figures).
This applet by Benjamin
Wardhaugh demonstrates the relationship between the string
lengths and the notes. The ratio ${(2)}^{1/12}$ is used to build
up the other intervals, so that each interval is a whole number
of semitones, and the ratio between its frequency and the
frequency of the lowest note in the scale is given by a power of
${(2)}^{1/12}$. For example the fifth is ${(2)}^{7/12}$.
Instrument tuners customarily use a logarithmic unit of measure,
the cent, where 1200 cents are equal to one octave, a frequency
ratio of 2/1, so that a cent is a 1200th root of 2. The table
below shows the Equal tempered, Pythagorean and Just systems of
tuning given in cents showing how many cents the note lies above
the starting C.
Fill in the
table.
|
C |
D |
E |
F |
G |
A |
B |
C |
| Equal tempered scale |
0 |
200 |
400 |
500 |
700 |
900 |
1100 |
1200 |
| Pythagorean scale |
0 |
|
|
|
702 |
|
|
1200 |
| Just intonation |
0 |
|
386 |
|
702
|
|
|
1200 |